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What is the final antiderivative of excos⁡xe^x \cos x?

The final antiderivative is ex(sin⁡x+cos⁡x)2+C\frac{e^x(\sin x + \cos x)}{2} + C. This result is obtained by solving the algebraic equation formed when the original integral reappears after two rounds of integration by parts.

Conditions

  • The function integrated is excos⁡xe^x \cos x.
  • The constant of integration CC is included.
  • The algebraic solution for the recurring integral is correct.

Reasoning, step by step

  1. Start with I=∫excos⁡xdxI = \int e^x \cos x dx.
  2. Apply integration by parts twice to get I=ex(sin⁡x−cos⁡x)2+⋯−II = \frac{e^x(\sin x - \cos x)}{2} + \dots - I (simplified conceptual step).
  3. Solve 2I=ex(sin⁡x+cos⁡x)2I = e^x(\sin x + \cos x) (conceptual intermediate).
  4. Divide by 2 and add CC.
  5. Result: ex(sin⁡x+cos⁡x)2+C\frac{e^x(\sin x + \cos x)}{2} + C.

Example

The script states: 'Moving it across gives the antiderivative eˣ(sin⁡x+cos⁡x\sin x+\cos x)/2+C.'

Common misconceptions

  • Omitting the division by 2.
  • Forgetting the constant CC.
  • Sign errors in the sine or cosine terms.

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